Three Dairy Production Numbers You Need to Look At
One of my favorite phrases that I hear is, “we are drowning in information while starving for wisdom.” I am sure everyone has felt this at some point in their life. Dairy farms are great at recording data and collecting information, however analysis of this information can be challenging. When evaluating data, these three production numbers should be looked at so as not to lose sight of the big picture.
Pounds of Solids: I am sure that you have read or heard someone talking about pounds of solids in an article or a presentation before. This is a calculation that is done to find out how many pounds of fat and protein the average cow is producing. When I ask producers about their milk components, they almost always tell me about their percentage of fat and protein. But is this really the best answer? Yes, a 4% fat and a 3.1% protein are good, but this answer tells a different story at 95 lbs. of milk compared to 85 lbs. of milk.
To calculate the percentage of fat and protein, simply add the two percentages together and multiple it by the milk production. For this example, take 4% fat plus 3.1% protein = 7.1%. Multiply 7.1% by 85 lbs. to get 6.035 lbs. of solids. 7.1% multiplied by 95 lbs. is 6.745 lbs. This is a massive difference both in production of the cow and the financial bottom line that cow is generating.
Why is this important? Producers get paid based on pounds of solids and pounds of protein—not the percentages of each. I often challenge producers to think about making decisions based on how many pounds of solids are shipped down their driveway instead of average pounds of milk production.
Pounds per stall: This number is valuable to help us understand the total farm efficiency and a way to increase the farm’s bottom line. For our example we will use a four-pen barn with 40 stalls in each pen. This would give us room for 160 cows at 100% stocking rate and 200 cows at 125% stocking rate. Since very few producers are at 100% stocking rate, let’s use the 200-cow example.
These cows are averaging 85 lbs. of milk/day. 200 cows multiplied by 85 lbs. of milk, divided by 160 stalls equals 106.25 lbs. of milk per stall.
We have worked with many of our producers to look at making their herd more efficient. We have done this by increasing their milk cow numbers by 10%. For this example, they would milk 20 more cows, or 5 more cows per pen. If we assume average milk production stays the same, you would calculate the new efficiency by taking 220 cows multiplied by 85 lbs. of milk and then divide by 160 stalls to get 116.88 lbs. of milk per stall. This means that average milk production can go down to 77.25 lbs. of milk and you will be shipping the same amount of milk as you were at 200 cows. We have experienced a very minimal drop in milk, if any at all, when we have implemented this thought process. Yes, management must increase and there are costs associated with feeding the cows, but your fixed daily costs do not go up and thus your return over feed cost is very profitable with these extra 20 cows.
This measurement of whole farm efficiency can also be very effective in evaluating percentage of your herd that is first lactation versus mature cows. A herd with a high percentage of cows in their first lactation (over 40% of the herd) is using stalls in their barn for cows producing 10-15 lbs. of milk less than mature cows. Therefore, their milk per stall measurement is less, resulting in the entire barn being less efficient.
Component efficiency: Component efficiency measures how efficient your cows are at producing their components. In other words, how many pounds of dry matter does it take to produce the pounds of fat and protein in the milk. Let’s use the same example for a cow producing 85 lbs. of milk with a 4% fat and 3.1% protein or, 6.035 lbs. of solids. If the average cow eats 57 lbs. of dry matter her component efficiency can be calculated by taking 6.035 lbs. of solids and dividing by 57 lbs. of dry matter, which equals an efficiency rate of 10.58%. Compared to a cow that may produce 87 lbs. of milk with a 4.2% fat and 3.2% protein or, 6.438 lbs. of solids, while eating 61 lbs. of dry matter or an efficiency rate of 10.55%. On paper you see the higher milk and higher components and think that herd is doing better but, in reality because of the higher intake, the 85 lb. herd is more efficient.
Now this doesn’t mean that the herd that is more efficient is more profitable. Many factors play into profitability like component pay price, percentage of the herd that is milking, feed costs, etc. This just means that there may need to be a discussion regarding ration efficiency and what forages and biproducts are being (or could be) fed to make the cows more efficient.
These three numbers are some of the many numbers we can calculate and track to help better understand your herd’s efficiency and profitability. Our goal is always to have healthy cows first and do what we can to help our producers increase their bottom line for their business.